geometry-and-trigonometry

[mystic grotto]

oh okay

ill check it now

thenks

[broken iron]

here remainder is -4

[teal flower]

There is a formula for $1 - \sin 2\theta$ in terms of $cos^2 \theta$ right?

[somber coyote]

NEONPerseus

[snow kite]

mfw trig seems so unintegrated with the rest of my math skills. whenever i have to use it in calc, it doesn't flow that naturally. should i just commit to memorizing all trig identities back to front or what?

[real wyvern]

Derive them ever day till you remember

That’s how I did them

[snow kite]

I'll give it a shot. Did the sine and cosine rule in the past, but i'll try the rest.

[lament smelt]

anyone have any resources on trig?

[upper karma]

Is the equation cosx = sin(sinx) solvable without the use of calculator or graph? If yes, how'd one go about it?

The unit circle, where the given coordinates are in the form (cosA,sinA), and A is your angle given there in radians

Oh 👍

[upper karma]

Unit Circle

[upper karma]

Try looking at the Trig book used by the USA Mathematical Olympiad Team. 😉

[floral shale]

Not really

I mean $x = \arccos(\sin(\sin(x)))$

[somber coyote]

Umbraleviathan

[floral shale]

Not sure what you wanna do with that

I mean you can try using Maclaurin series

Not fun

Alright @upper karma I'm assuming you don't know what Taylor series are but

You can approximate a value for x that is between 1 and 3

$$1-\frac{x^2}{2!} = \left(x-\frac{x^3}{3!}\right) - \frac{\left(x-\frac{x^3}{3!}\right)^3}{3!}$$

[somber coyote]

Umbraleviathan

[limber frigate]

ok so I'm pretty sure this is technically algebra but this is in my geometry class so
How do I simplify the sqrt of 169?

I tried sqrt of 64 times sqrt of 2 but that was wrong.
64 canceled into 8 so 8 sqrts of 2.

[upper karma]

👍

[upper karma]

oh dear, can somebody help me out? I still don't understand the tangent line function, and where it comes from in the unit circle I see that Y coordinates represent the sin(theta) and X represents the cos(theta) what does tan(theta) represent??

[upper karma]

Guys does graph transformations come under this channel?

[nocturne remnant]

Probably would be more fit for #precalculus I think

[floral shale]

169 is one of your perfect squares that you should've memorized

[upper karma]

Yeah. 13^2

[limber frigate]

I only know them up to 10

[floral shale]

You need to know up to 15 and your 10s

[last cypress]

I am in a Honors Algebra II/Trig class that essentially combines the two classes, and I need to buy a TI-Nspire CX. Would you guys reccomend the CAS version or no?

[upper karma]

,rotate

[upper karma]

What we have to do in this

[smoky jetty]

honestly, I've been here as well before and failed my Trigo exam. But something I learned was to understand the concepts and the reasons why those become such things etc. After re-studying my Trigo book, I realized that I'm using some formulas effortlessly without the need to memorize them a lot. It's just me giving an opinion on how to undertake it, take it with a grain of salt if u will

but it takes a lot of time for me, and some things can be forgotten after going to other lessons. Hopefully, I can just look back and read a bit and remember the concepts yet again when I'll need them, lmao poor me

radian system and some functions/ratios? i recognize those, lol

[lament smelt]

i just understood the radian system in 1 wikipedia animation

[upper quarry]

I just need a study partner for trig

[copper trench]

who can help me with 5 geometry questions

[brittle sparrow]

can someone explain to me the difference between solving for sin a cos a tan a and sin c cos c tan c

[wild path]

Please help solve this challenging geometry questions. Thanks!!!

[spark panther]

takin trig for the 2nd time... anyone good at helping teach?

[snow kite]

Yeah I miss out on that a lot too. Why and how did those identities come about? What problems were they trying to solve? But I did some looking on math stack exchange and saw that all those identities could be derived with demoivre's theorem, but it involves complex numbers.

[smoky jetty]

there's no difference with the concept of finding the answer, it's just that you'll solve for the functions for the other angle, which is C

[smoky jetty]

identities be like, lmao

[smoky jetty]

basically, the value of the ratio will (can) be different, but the way u find the ratio for sin A, cos A, tan A, will be the same for sin C, etc. Nerd folks can prob explain better, lol

[eager bolt]

lol

[balmy sun]

I recently came across this really old trigonometry textbook which uses interpolation in order to find the value of a function of an angle. In this case I'm wondering where the 60' value (see above picture) comes from which is used in the estimate ?

for reference the book is Lymann Kells - Plane and Spherical Trigonometry

[dark sparrow]

you're doing linear interpolation between the values of sine at the two nearest integer-degree angles to yours

they are one degree, or sixty arcminutes, apart

[grave pond]

You mean he area of the green square?

You can find its side length using the Pythagorean theorem. (If you draw the configuration on graph paper with two squares per unit distance, everything will align neatly with grid points).

[dark sparrow]

is coordinate-bashing allowed? i have a feeling this problem might yield pretty easily to it

...or that, why not

[balmy sun]

@dark sparrow Thanks, that clears it up. So 1 arcminute = 1/60 of a degree. Degrees are further divided into arcminutes. I suppose this has its use in astronomy and the fact that 360 is easily divisible into 60?

[dark sparrow]

minutes and seconds of arc are still used in navigation to communicate lat and long

they are a historical remnant of formerly common sexagesimal subdivisions

[balmy sun]

yeah, from the babylonians

Glad to hear its still relevant. I skipped the whole section on slide rulers, that felt a bit more outdated and who has access to one anyways? haha

[dreamy crown]

47 is the answer right?

[dreamy crown]

<@&286206848099549185>

[grave pond]

How do you get that answer?

[dreamy crown]

Cuz half of 94 is 47

[grave pond]

But then you're missing the triangle with the 75° angle in it.

[dreamy crown]

why does that matter

94 is still a central able

Angle

Without using it

Ok the answer is 62 then

Based off a estimate

That’s the only reasonable one

[grave pond]

94 is indeed a central angle, but the peripheral angle you're asked to find doesn't span the ends of the 94° arc...

The central angle that matters is 94° plus the arc adjacent to the 75° one.

[dreamy crown]

So???

What does this mean

How do I get a answer based off this

[grave pond]

Lets give the other points some names.

The central angle theorem tells us that angle AMB is 47°.

But the angle we're looking for is the sum of angle AMB and BMC.

So you'll need to reason about triangle OBC to find out the length of the arc BC and then use the central angle theorem again to get angle BMC.

[dreamy crown]

can u just tell me the answer please

I’ll figure out why its that on my own

I think its 62 so far

[grave pond]

We don't give out answers.

[dreamy crown]

So 62 is the final answer

thanks

[stone tulip]

The face of a sphere is 2 dimensional?

[grave pond]

Yes.

(But "face" is an unusual word to pick in that context. We'd say "surface" or "boundary" instead, to distinguish between a solid shape and its outside).

[undone moat]

Wait.

If you join A and C it would hypothetically divide the triangle into 2 pieces

one of the parts worth 2/3 &1/3 respectively

going on that line I got 47 degrees

I dunno whether it is correct

but I tried

[grave pond]

Which triangle would be divided into two pieces?

[undone moat]

Well by joing a and c

I would make a triangle

which would seem divided into two

smaller triangles

[grave pond]

A triangle needs three points, "A and C" are only two.

[undone moat]

o is another one

angle aoc is the largest angle of that triangle by observation

[grave pond]

I think what you're trying to say that angle AOC = angle AOB + angle BOC?

[undone moat]

no

I am saying that I made another triangle, triangle aoc

and since that triangle seems to be 1/3 of amc

with that triangle looking like an isosceles

nvrmind

I think I got the question wrong

[dense canyon]

Can someone help me with 39

[floral shale]

All but e can happen

Imagine if the planes were parallel

And then imagine two parallel planes and one normal plane

[small glacier]

Someone explain to me what the answer to this is

For 21

[upper karma]

have you tried anything?

[small glacier]

On a different thing of paper yes

[upper karma]

can i see it?

[small glacier]

And I’ve asked my parents

[small glacier]

That paper is gone I’ve done the whole packet but that and 1 other question

[upper karma]

uh alright i guess

then take another paper, have you tried drawing it?

geometry problems require usually a good drawing to visualise the problem

[small glacier]

No teacher haven’t taught us geometry and I’m in a algebra class so idk y she gave us geometry

[upper karma]

this is pretty much algebra but you may find having a drawing easier to solve it

take a paper first

do you have it?

[small glacier]

Yes

[upper karma]

alright then first let's draw this

[small glacier]

K

What am I meant to draw

[small glacier]

Would a call be easier

[upper karma]

something like this, that it adjusts to what the problem is saying

[upper karma]

not for me right now

when you have that draw let me know

[small glacier]

K

I’m terrible at drawing so this is gonna be fun

But next step

[upper karma]

have you read the problem and understood it?

the problem is talking about one angle which for us is the green one

so you can label it as x

[small glacier]

Ok

It’s starting to make sense kinda

[upper karma]

right so now the set up is done, the key sentence is:
"the measure of the supplement of an angle is 10° more than twice the measure of the complement of the angle"

[small glacier]

Yea

[upper karma]

do you know what the complement of an angle is?

[small glacier]

No I haven’t been taught this

[upper karma]

would you be able to locate it on this drawing? (which color)

[upper karma]

sure but the question at the end tells us that it is the angle that adds up 90° with the original angle

does that make it clearer?

[small glacier]

Yes

[upper karma]

so then which color is it?

out of these three?

remember we assigned green to be x, the angle the problem is talking about

so it's complement which color is it?

[small glacier]

Y is the purple like that

Would it be the 2X and the green plus 10 percent

[upper karma]

because it starts from that horizontal line to the diagonal line

[upper karma]

what?

[small glacier]

Cuz it says the measure of the supplement of the angle is 10 more then twice the measurement of the complement of the angle

[upper karma]

we aren't there yet, i'm asking you which color is the complement of the green

[upper karma]

and no that's not really correct

x isn't the complement nor the suplement angle, let's go step by step

[small glacier]

So green isn’t the 10 percent

[upper karma]

no, they say "10° more"

and you still aren't answering my question

[small glacier]

Idk what green would be cuz u said it ain’t a complementary or supplement angle and it ain’t the 10 more

[upper karma]

that's not my question, green is the angle the problem is talking about, it isn't the complementary nor the suplementary.

the problem is talking about the complementary and suplementary of THAT green angle

but not about the green angle

[upper karma]

can you point which angle is the complement of the green angle in this pic?

[upper karma]

remember the definition of complement i said here

[small glacier]

Would the complement be the blue and purple and the supplementary be just the big purple angle

[upper karma]

correct, yes

so now

what's the measure of the suplement of x?

it's given in the problem

[small glacier]

It’s 180-x

[upper karma]

great

and complement 90-x

do you agree?

[small glacier]

Yes

[upper karma]

so finally

with that sentence, are you able to translate it into an equation?

[small glacier]

Yes?

But at the same time no?

But idk what x equals

Cuz it says write an solve

[upper karma]

you have to find what x equals with an equation

[small glacier]

I think

[upper karma]

wdym by this

your objective is to write an equation like 4x+12=2(x-100) and then solve x with algebra

[small glacier]

Cuz it says write and solve an equation to find the measure of the angle

Oh ok

[upper karma]

yes, but first you have to draw the equation

[upper karma]

and you write the equation with this sentence

[small glacier]

Wait draw u mean write

[upper karma]

what? by write i mean to make up the equation

no drawings

[small glacier]

K

So supplement equals complementary times 2 plus 10

[upper karma]

exactly

[small glacier]

But how would I write it as equation cuz I can’t use words

[upper karma]

well you told me the measure of the complement of x and suplement here

[upper karma]

and here

[small glacier]

So 180-x? Or do I have to add something else

Yo u still there

@upper karma

[small glacier]

Nvm I got it

[slate apex]

Are triangles, squares, and hexagons the only basic polygons that can completely tile a surface? (i.e. are there any numbers besides 3, 4, and 6 that can do this?)

[grave pond]

They're the only regular polygons that can tile the plane.

The corner angles of every other regular polygon do not divide 360°.

[warm tulip]

line s, 48

[stray oak]

hello is there any good full youtube video to learn geometry? I took it on aleks 10 years ago, same with algebra 1 and 2, but other than that i am relearning

greenemath i am using for algebra 1 and 2 but don't know what to use for geometry, i know khan academy but i was looking for more something all together

[undone moat]

https://youtu.be/302eJ3TzJQU

It is an introduction if that is what you want.

[slate apex]

Thank you! I didn't see your reply yesterday

[upper karma]

I'm doing my friends math homework and I was wondering how you right a hypothesis and conclusion for the problem if 3x -7 = 32 then x = 13

Im not in the geometry class

So idk

Or what would be the right channel to ask that again

Nvm found ot

[sick breach]

Anyone know how to find x?

[dark sparrow]

let O be the point of intersection of WY and XZ

then you have four right triangles

WOX, XOY, YOZ, ZOW

[sick breach]

yup, i've got that far, including finding the value of x

i can find the value of every angle, except the angles in XOY, where i only know that the missing angles sum to 90

[smoky jetty]

oh wow, i honestly still use it for some problem. sometimes converting D-M-S to Decimal degree measurement vice versa helps me do a problem more digestible.

[dark sparrow]

oh wait yeah you're right

think you might be able to do some trigonometry bullshit tho

let OX = 1 and express OW, OZ and OY in terms of the known angles

then you will get the value of tan(X)

[sick breach]

right, i suppose that would work

ty

[teal lantern]

hey i need

help

anybody here?

[smoky jetty]

since the ZWO triangle is a special right triangle case (30-60-90) triangle, then the value of the shorter side is a. While the longer side is a*square root of 3, and the hypotenuse is a(2). U can find out why they're like that using the Pythagorean theorem or trigonometric table. Then you'll have enough side values for both ZOY and WOX triangles, then use those 2 sides to find the other side and the angles of the XOY triangle using the Pythagorean theorem and solution of a right triangle. these are based on what learned from self-studying, so I'm not 100% certain, but I'm pretty confident that it's the only way, lol

[teal lantern]

questions from multiple angles

from grade 10

[smoky jetty]

btw kindly tell me if I responded too specifically that may disobey the rules here

[teal lantern]

@smoky jetty

[smoky jetty]

sorry dude, im not familiar with that identity

[smoky jetty]

basically you'll already have 1 given side each for WOX and ZOY triangle since WOZ triangle is a special right triangle case and it already has values. So you'll use the side and the given angle of WOX and ZOY to find the other side, of those 2 triangles, which are also the sides of XOY triangle, since they're composite figures. You may also find the other complementary angles of those triangles if necessary. been a while since I did such problems, so im a bit unfamiliar with the procedures, sigh

[sick breach]

yh, solution sounds all good tho

[nocturne remnant]

I’ve read this book before… i think the author intended for the reader to not be able to do (b) until they learn about cyclic quadrilaterals in the next section

[smoky jetty]

dang, is that geometry 9th grade? then im not sure with my response @sick breach . But I learned the pythagorean and solution of right triangle mostly by self studying ater 8th grade, and it seems like it's doable with those

[sick breach]

is it solvable using no trig/pythagoras then?

[nocturne remnant]

Yeah technically

But why wild you do that

Would

[sick breach]

learning these methods for competitive maths

[nocturne remnant]

I know… but why not follow the intention of the author?

[smoky jetty]

what's the intention of the author? lol

[sick breach]

isn't the intention of the author not to use pythagoras and trig?

i was unsure if it could be done, i kinda just stared at the question for a while

[smoky jetty]

oh, welp thats beyond my pay check

[smoky jetty]

it's doable with trigo stuff imo, but yeah im not 100% sure

[nocturne remnant]

This is from a few pages later

Lol

[smoky jetty]

lol now I have no clue

questioning myself where the ABCD came from, lmao

[sick breach]

right, so we can use the fact that angles subtended by the same arc have the same measure to say that it is cyclic

then the problem become's trivial

[nocturne remnant]

Yep

[sick breach]

i had that idea before, but i created a construction that convinced me that the converse of that theorem isn't necessarily true

[nocturne remnant]

As in?

You mean the two subtended angles lie on opposite sides* of the chord?

[sick breach]

consider the same quadrilateral as above

we have the two similar triangles ZOY and WOX

[nocturne remnant]

Ye

[sick breach]

now if it is cyclic, then we can circumscribe a circle around every vertex

so lets do that

oh shii

i just realised my mistake

the circumcircle changes as you change the quadilateral

welp thanks nonetheless

[smoky jetty]

quit geometry and embrace trigonometry lesson, lmao jk

[upper karma]

what is the difference between period and wavelength of a wave?

[sick breach]

if you imagine a wave moving through space, time period is the time it takes for the wave to move one full wave, so like it it's in the same position as it was when it started moving

the wavelength is the physical length moved during that time, the length between two equivalent points

[upper karma]

ohhhh

[sick breach]

note that one is a time, and one is a length

[upper karma]

phase difference?

[sick breach]

think of translating sin waves

[upper karma]

yeah

the distance between shifted crests of waveS?

[sick breach]

it's the value in radians you would have to shift by

[upper karma]

oh

[sick breach]

the distance is the path difference

[upper karma]

okay

[sick breach]

phase difference is more a measurement of how out of sync it is as a measurement of an angle

harder to think about, but i like to think of sin waves and translating them

[upper karma]

so phase difference is the value of radians you need to shift it to get the waves in sync?

[sick breach]

yup

[upper karma]

last thing

what is phase angle

[sick breach]

i've never come across the term phase angle, but i'd assume it's the same as phase difference, since it's an angle

take that with a grain of salt though

[upper karma]

alr

[smoky jetty]

period is the interval through which the value repeats, and wavelength is pretty much what harry said

[upper karma]

time interval for whatever value repeats, like a crest to another?

[smoky jetty]

not sure with what u said, but is crest like the maximum amplitude?

i often think of period as the interval of the function where the curve starts to repeat, from the origin

[upper karma]

yeah

[vapid jolt]

I have a bit of trouble when it comes to construct shapes using the "Analysis, Construction, Proof, Discussion" pattern. May someone help me on this?

[floral moon]

The strength of Earth's magnetic field varies with the depth below the surface. The strength at depth z and time t can sometimes be approximated using the following damped sine wave.
$S = A_o e^{-az} \sin({kt - \alpha z})$ where $A_o$, $\alpha$, and k are constants
At what depth is the amplitude of the wave one-half the amplitude of the surface strength?

[somber coyote]

Chrovo

[floral shale]

Seems like preuniversity stuff to me ngl

[grizzled wigeon]

Hi

Does this mean I have to copy all of these angles using a compass?

[lyric hedge]

I am confused on how they got 6.89m as the answer

I think I'm get Ha and Hb length wrong

I did
d * tan(θa) = Ha
Ha = 40.27
d * tan(θb) = Hb
Hb = 26.37

and then subtracted Hb from Ha but got 13.9m and the answer is 6.89m

[lyric hedge]

brhuhh

my calculator wasn't in degree mode.......

I've been wondering why I was getting the wrong answers for all my questions for like an hour now

[lyric hedge]

"Two racing boats set out from the same dock and speed away at the same constant speed of 97 km/h for half an hour (0.500 h), the blue boat headed 22.0° south of west, and the green boat headed 38.0° south of west."
Can anyone help me through this problem

I just need to know how to approach it to get the formula to solve it

because I'm not really sure how to approach this one

[smoky jetty]

eek

[smoky jetty]

does the problem not have an illustration?

[smoky jetty]

I think this has something to do with the "Bearing of a point" trigo thing

try looking that lesson up on google or in any resources u got.

pretty much a right triangle problem as well

[shadow grove]

<@&286206848099549185> can someone help me create a formula for a non linear pattern

Please @ me

[buoyant panther]

shut up nerds

[upper karma]

Please help, I'm trying to solve x for alpha and cant find a way of getting to a quadratic equation or smth

[silent plank]

consider harmonic trig identities to express the left side with a single sine function

[hidden crescent]

suppose you have a line and point outside the line

what do you call the shortest distance from the point to the line

ik it makes a right angle with the line

[hushed shore]

a perpendicular line?

[silver silo]

perpendicular distance?

if the line is ax + by + c and your point, (x1, y1), the general formula for that distance is abs(ax1+by1+c)/sqrt(a^2+b^2).

[visual coyote]

hi, was just reading through some proofs of euler's formula (for polyhedra), then came across this one and i have no idea where the highlighted yellow part comes from - any ideas? thanks

[empty ferry]

hi guys im in 7th grade rn and taking alg 1 rn next year in 8th grade im looking to skip highschool geometry and go to alg 2. My teacher is on board with this plan and is giving me some of the stuff that you learn in geometry but i was just wondering if someone could tell me difficult concepts to work on for geometry i can work on at home ty

[velvet bramble]

How would you find coordinates of last parallelogram

[grave pond]

It looks ike they're subtracting some multiple of (c,d) from (a,b) such that the y-coordinate becomes 0.

[glacial inlet]

for laws of sines how can you tell if theres 2 triangles or 1 triangle

I know if sin B > 1, then there is no triangle, and if sin B = 1, then its a right triangle

but i've been confused on the other 2 cases

[glacial inlet]

nvm I get it now

[shy stratus]

Does anyone know how to solve the area of the shaded part?

[grave pond]

Divide it into congruent quarters by a horizontal and a vertical line.

Each quarter is then a sector or a circle minus two triangles.

Then you only need to find some angles and lengths.

[shy stratus]

Is there a way to do this for curved lines?

[grave pond]

The lines are curved.

As in, they are circular arcs.

Wait, am I misunderstanding what you're asking?

[shy stratus]

Um so here I’m saying like these lines here

For example

[grave pond]

I'm pretty sure they're part of circles.

That's the assumption behind my hint.

[shy stratus]

Wait to clear my confusion when you said minus two triangles are you referring to these

[grave pond]

No.

[shy stratus]

Oh

[grave pond]

These two.

[shy stratus]

Woah

And from here would you be able to use trigonometry

[rose tendon]

i know that the sine and cosine of a rational multiple of pi is always algebraic but would the opposite be true where a non rational multiple of pi output a transcendental?

[grave pond]

No.

(The sines and cosines of rational multiples of pi always lie in a cyclotomic field, but those fields don't give all of the algebraic numbers).

For one particular example, $\arcsin\Bigl(\dfrac{\sqrt[3]2}{2}\Bigr)$ is not a rational multiple of $\pi$.

[somber coyote]

Troposphere

[grave pond]

... I think. The argument I had in mind for that doesn't hold together when I try to write it down. :-/

[stuck sleet]

anyone knows a good plane and solid geometry book?

[proper rune]

<@&286206848099549185>

[sly raptor]

Well what do you know about equilateral triangles

All angles will be 60 degrees and all sides are the same right

@iron blade

We also know that the side of an equilateral triangle is equal to (2 * altitude) / sqrt(3)

So, 24 / sqrt(3)

Which simplifies to 8sqrt(3)

[lyric hedge]

Will the angle always be in spot 2 or can it be in spot 1

[lyric hedge]

for example if the problem says the angle is 30 degrees north of east

wouldnt I draw the circle inn spot 1

[grave pond]

"30 degrees north of east" means that the angle beween a line due east and the direction you're thinking of is 30°.

So you should draw the arc that symbolizes angle measurements between those two directions.

Otherwise it would look like you're talking about a direction 30° east of north instead.

[lyric hedge]

for this I need to find the hyp for bicyclist 1 and since they have the same start and end I use that hyp for bicyclist 2 to find the missing side which should be the answer but I'm getting it wrong

[grave pond]

What is a "hyp"?

[proven pilot]

Hypotenuse?

[grave pond]

Hmm, perhaps. But "the hypotenuse of a bicyclist" doesn't seem to make a lot of sense still.

[proven pilot]

Oh. You are solving his question

[grave pond]

Why, perish the thought! I was aiming to help him solve it himself. But he's not giving us a lot to work with.

[smoky jetty]

as far is I know, the reference angle to quadrant one (perhaps to all quadrants?) is always relating to the x axis

[smoky jetty]

yeah great point

[smoky jetty]

oh bearing of a point tho, i think it depends on the statement given jg

[grave pond]

The trouble here is in pure mathematics we usually measure directions counterclockwise from "due right", whereas in navigation we measure directions clockwise from "due north". Since the exercise tries to disguise a problem in pure coordinate geometry as if it were about navigation, it needs to specify which of the conventions it wants the result given in.

[autumn ridge]

I'm in australia so uh. y =mx+c is the rule for a straight line. so say uh
y=6x+8 The gradient (m = 6) but what if it was y=3-7x?

would M = -7?

and would i have to add the x or just leave it like that.

I'm doing revision for my exam tomorrow.

I'm guessing this is part of Geometry.

I'm only in year 9 so.

[grave pond]

Yes, -7.
I don't know what you would achieve by adding x.

[lyric hedge]

Hypotenuse

[lyric hedge]

I already have the context down I was wondering if my thought of process was correct

The hypotenuse as in the distance the bicyclist traveled

[lyric hedge]

It keeps saying this is wrong but idk why

I use 97 as the hyp and then got the adj and opp, and then subtracted them from each other to get the difference and that difference should be the answer right?

me and my friend doing the same process but getting it wrong so is my process wrong

[lyric hedge]

I found the answer to this one btw

[lyric hedge]

I just need help with this one

The question seems easy but I’m getting it wrong

[iron trench]

trigo is my weakest

[nocturne remnant]

i'd say this is more coordinate geometry rather than trigo

which ones do you need help with?

[iron trench]

numbers 3,4,5, and 6

[proven pilot]

What is 2+2

[upper karma]

i hate trigo

i just started taking it

[next mantle]

https://tenor.com/view/perspective-different-viewpoints-gif-22252829

now you wont hate it

[slate mauve]

https://cdn.discordapp.com/attachments/792943252889141281/1016799118691155968/IMG_3966.jpg

I understand that I must plot coordinates that must consist of real numbers (Hence the 2 Lined R squared) but I cannot understand the x^2+y^2=0 part. Usually when I am assigned questions of these there can be a clear answer on the slope and such but this one I am unable to understand.

[grave pond]

There is exactly one point that satisfies that condition.

[slate mauve]

the origin?

[grave pond]

Yes.

[slate mauve]

ty

i honestly wasnt too sure'

[grave pond]

For example, if x is not 0, then x² > 0, and then y² = 0 - x² would need to be negative, which is impossible for a square.

[slate mauve]

i did -y^2 = x^2 so ye

did not work well

[grave pond]

That works excellently -- all you need is the courage to draw the right conclusion from it.

[slate mauve]

so i just used intercept formula

ye

[exotic yarrow]

Note that the midpoint of the diameter is the center

and the radius is the distance from the center to an endpoint of the diameter

[fading veldt]

must two triangles with the same perimeter be congruent if the triangles have one angle measure in common?

[naive obsidian]

Help

[grave pond]

That's a horribly buggy problem.

[empty ferry]

Which one

[naive obsidian]

E

The one on the bottom left

[empty ferry]

The squares area is

[grave pond]

The middle left diagram seems to be claiming that 4-6-8 and 3-6-9 triangles are right!
The bottom left diagram also has a purported 3-6-9 right triangle sticking out from the rectangle.
Digram N in the bottom center claims there are 4-7-9 and 5-6-10 right triangles.

[empty ferry]

56

The lower square is

18

[grave pond]

And Y at the bottom right needs a 5-10-11 triangle.

Every single diagram with a printed number next to a diagonal edge is impossible!

[empty ferry]

The triangle is 9

Add it up

[naive obsidian]

Ok thank you guys

[empty ferry]

Thats how they make these pre alg things

Doesn’t work with geometry lmao

[grave pond]

Hmm, no I suppose the D in the top center is possible if we assume only one of the diagonal edges has length 7. But its area should be 64 (8×6 + ½·8×4).

[naive obsidian]

Isnt it supposed to be 68 squared

[grave pond]

How would you get 68?

Note that 10 is the dashed height, not the width of the shape.

[naive obsidian]

Half of 10 is 5 if u cut it to make a rectangle and triangle

[grave pond]

The height of the rectangle at the bottom is stated to be 6 -- nobody says that has to be half of the total height.

[formal hatch]

i need help

[lyric hedge]

just post the problem and say what you need help with

and someone will help you when they see it

https://dontasktoask.com/

[lethal canopy]

Can someone explain to me what bissectors are?

Also, what is the law of cosines and sines and tangents?

[quaint osprey]

I only picked up s/c/t from an online course, but maybe I can help - do you mean basically how they're derived?

[grave pond]

He's probably referring to https://en.wikipedia.org/wiki/Law_of_cosines and https://en.wikipedia.org/wiki/Law_of_sines.
I never knew there's also a https://en.wikipedia.org/wiki/Law_of_tangents, but Wikipedia does list one.

[smoky jetty]

oohh law of tangents? didnt know that

[grave pond]

What are you supposed to do with that? Solve for theta? Simplify?

[next mantle]

prove both sides are equal?

[snow kite]

it looks like proving the identity given that the next problem has multiple variables

[snow kite]

start from either side, and use identities to get it into the form on the opposite side

[rugged pebble]

If the Penrose tiling can tile 2D space without repeating itself, is there an equivalent that can tile the 1D space without repeats as well?

My instinct is no, but even then I don't know what changed between 1D and 2D to allow the Penrose tiling to happen

[random garden]

Hi, I'd like to figure out what sections of specific longitudes / latitudes are visible for a sphere from a given viewpoint. Any suggestions? (in three dimensions)

[rugged pebble]

You can use triangulation and the radius of the sphere to derive the horizon at any given height

This is actually how one of the first meseaurments of the earth's radius was taken, because you know your height and the distance to the horizon, you can work backwards to find the radius of the sphere

[cerulean wasp]

Can someone help me with my geometry hw

[floral shale]

Just post it

[dark sparrow]

this person left the server

[alpine tiger]

∠abc + ∠cde = ∠nop + ∠cde does any one know what the proof for this is

[dark sparrow]

there is no proof, because we do not know what points a, b, c, d, e, n, o and p are

(also presumably these are all meant to be uppercase...)

@alpine tiger

[upper karma]

can I see a picture of the points?

[upper karma]

well, it doesn't really matter. just recognize both sides have a like term, and the sides are equal. therefore the unlike terms also have to be equal.
a + b = c + a

[upper karma]

you might be better off asking this in #proofs-and-logic since you don't provide an image

so it isn't really a trig question

[silent plank]

i mean the question is then reduced to why
∠abc is equal to <nop
and we have no idea since we have no idea what those are
without the full problem

[upper karma]

i was trying to work with the fact that the middle letter is the point at which the angle is found between line segments to the other 2 letters. but none of the letters in the terms are the same except for the identical angles

apart from the left-term

[keen yoke]

Oooh

Been experimenting with matrices and its relation to 3-dimensional tranformations

and it is so handy to use them

[dark sparrow]

what kind of problems are you solving?

trigonometric simplification? identity verification? equations?

[silver stump]

Hello can anyone give me a hint for this question?

[grave pond]

Is angle ADC supposed to be right?

[silver stump]

nope

[grave pond]

Hmm, one thing I can think of is moving triangle ACD so it shares the 30° degree angle and the equal sides with BCD, becoming DBM here.
Then we know that angle CBD + angle DMB = 180°, implying that angle CDM = 30°.
Then ... oops, we've actually recreated exactly the same problem at a smaller scale.

But that may actually be progress, because it tells us that if the solution is unique, then angle DCB is also x.

[silver stump]

i am super confused

[grave pond]

But ... that leads to the conclusion that x=60°, which doesn't work if we actually draw the configuration. So either I've made a mistake somewhere, or the solution isn't determined by the givens, or the whole thing is impossible.

It's not actually obvious that the whole thing is possible. Suppose we start by selecting points B and C in the plane, and draw he 30° angle at B. Now we place the tip of another 30° wedge at C and begin slowly turn it, producing possible positions for A and D as it turns. Is it clear that there's any orientation of the second wedge that will make CA have the same length as BD? Both of the extreme orientations of the wedge makes CA much longer than BD.

[silver stump]

hmm ill ask our teacher

[grave pond]

Even if there is an angle where CA is shorter than BD, that would just mean that there are at least two intermediate angles where CA=BD.

[silver stump]

ok thanks for the help!

[grave pond]

In summary, what I feel reasonably sure of is
a) if x is a solution, then 120°-x is also a solution;
b) x=60° is not a solution.

Wait, in the x=60° case we do get DB = 1.5CA.

So there is in fact a solution with 0<x<60°, and another solution with 60°<x<120°.

[silver stump]

so it has 2 answers?

[grave pond]

I did some coordinate bashing and threw the resulting equation at Wolfram Alpha which said that x=24° and x=96° are solutions.

[grave pond]

Here are the two winning configurations, to scale.

[silver stump]

oh my god thank you so much

Just a quick question. are these 2 the only possible answers?

and what is the process of getting the answers

[grave pond]

It looks like those are the two only answers.

I got them by punching a horrible trigonometric equation into a graphing calculator and asking for solutions.

[silver stump]

oh dear lord

[grave pond]

Still trying to figure out a way to prove they are solutions without trusting a calculator.

In each case, the correctness comes down (via a few applications of the law of sines) to checking that sin(30°)sin(126°) = sin(96°)sin(24°) for the upper configuration, and sin(30°)sin(54°) = sin(24°)sin(96°) for the lower configuration. This is of course the same calculation for both cases, since sin(126°)=sin(54°).

[silver stump]

there has to be a way of solving this problem without using sins and cosines because im in 7th grade

[grave pond]

Oh wow.

I have absolutely no idea what you could be expected to do about this problem with 7th-grade tools.

[silver stump]

Lmfao that was why i was so lost

i know what sins and cosines are and like the rules related to them but we havent learnt them in school yet

anyways thanks for the help mate!

[visual gale]

woah this convo lasted long

[silver stump]

I think i have something

The left triangle is similar 2 the big one in both cases

The only thing that differs the 2 is if either the buttom or the top side is the biggest one

In one the buttom left angle is less than the top on in the left triangle

And in the other one its the other way around

Does that make sense?

The small and big triangle share a side so we just have to determine if the shared side

Is the bigger or smaller one compared to the the buttom side in the small triangle

Thats why there is 2 cases

Does that make sense or is it 4 am and im just spitting random stuff lmao

[grave pond]

This is a very interesting observation which had completely escaped me.

[silver stump]

So we know all angles are the same and also we have the ratio of 2 sides

Let me try to draw it

[grave pond]

Yeah, I'm getting a bit lost in your verbal explanations.

[silver stump]

Poor english + geometry = death

Give me a sec

[grave pond]

(I was writing on a horrible solution full of trigonometry and algebra that would at least eliminate the dependence on calculators ...)

[silver stump]

Ignore the persian writings

So first we have to flip the triangle

So that its in the correct way compared to the big one

We have an angle so we need 2

To get those 2 we have to determine 1 thing

Either ac > ad or ad > ac

[grave pond]

Okay.

[silver stump]

And its all ratio from there

Or at least i believe so

[grave pond]

Hmm, "all ratio" sounds too easy here; I don't see how that would produce those particular solution angles.

[silver stump]

It sounds much more dumber after i wrote it down hmm

Quick question

Is it possible to find the ratio of rhe 2 angles using the ratio of the 2 sides in front of those angles?

[grave pond]

Almost, but you need trigonometry. The law of sines sas that the ratio of two sides in a triangle is the same as the ratio of the sines of the two angles opposite those sides.

[silver stump]

Ahh

So we cant just say ab/ac = ac/ ad

[grave pond]

Ah yes, we can say that because of the similar triangles.

[silver stump]

So then

It is possible without sins and cosine

S?

Ill try it

[grave pond]

Hmm, you can probably conclude that D must divide AB in the golden ratio.

[silver stump]

"Golden ratio"?

[grave pond]

https://en.wikipedia.org/wiki/Golden_ratio -- a famous geometric proportion which occurs (among other places) in pentagons and pentagrams. The difference between our two solutions 24° and 96­° is just 1/5 of 360°, so it's not unexpected that the golden ratio would show up.

[silver stump]

Hmm might not

Because the triangle is fliped

[grave pond]

For what it's worth, here's the horrible solution I've been able to cook up. Still pretty far above 7th grade, but at least doesn't depend on trusting a calculuator.
By angle sums in triangles we can derive these angles:

Now use the law of sines to write |DB| and |AC| as functions of |BC|, giving

|BC|·sin(120°-x)/sin(30°+x) = |BC|·sin(30°)/sin(x)
Cancel the |BC| and multiply across to get
sin(120°-x)sin(x) = sin(30°)sin(30°+x)
Use trigonometric product identities to rewrite the two sides to
½[cos(120°-2x) - cos(120)] = ½[cos(x)-cos(x+60°)]
Cancel the ½s and use the general identity cos(x) - cos(x+60°) = cos(x-60°) to arrive at
cos(120°-2x) - cos(120) = cos(x-60°)
cos(-2y) + 1/2 = cos(y)
where y = x-60° and finally
cos(y) - cos(2y) = 1/2
This is satisfied by y=±36° (which can be argued by considering coordinates in a regular pentagon with the origin in the middle of a horizontal side), which corresponds to the solutions x=24° and x=96°.

y=±108° also solves the last equation, but that would lead to geometrically impossible configurations with some of the angles being negative.

[silver stump]

I see

Thank you

Ill try to find a solution with 7th grade tools

[grave pond]

Good luck. There's probably something simple I'm overlooking ...

[wet ferry]

how is this a 7th grade problem

[upper karma]

it is

MEOW

[hot turtle]

24th

[hoary nymph]

https://cdn.discordapp.com/attachments/917359060964225044/1018186026142945280/unknown.png

does this triangle even exist? because in from triangle adb as all sides are given using cosine rule we find that portion of angle a which is arccos(5/6) using this info in triangle adm we try to find dm but if you use the cosine rule then dm turns out to be an imaginary number

[silver stump]

@grave pond hi sorry to bother again. We also have a shared altitude in both triangles. Hab = Had
Is that somehow usefull?

[pulsar grove]

So how do I learn this

Shapes and shi

Thought that was like pre k

[vital ginkgo]

Which book should I start with to learn geometry?

[worthy sapphire]

geometrically, let the pentagon be A1A2A3A4A5 and let the sum be S and let the side length be t. then 2S = (cycsum of a1^2 + a2^2) = (cycsum of t^2 + 2a1a2 cos(A1OA2)). now (cycsum of a1a2 cos(A1OA2)) can be expressed in terms of the area of the pentagon A1A2A3A4A5 using the sine area formula.

(motivation for this is letting α be a 2nd root of unity, then use pythagoras as a special case of cosine rule)

[hot turtle]

26th

??

[worthy sapphire]

i'm talking about question 24? wdym

[hot turtle]

I mean i also want 26th soln

Please

[worthy sapphire]

the question is cut off in the image

[hot turtle]

Sorry mate didnt notice

[worthy sapphire]

uwu rotate

[hot turtle]

Is the image full?

[worthy sapphire]

i cut it off myself

the rest of the image shows the rest of the solution but i just want to give the first steps

[smoky jetty]

jeez, i never thought la of sines can be this wild (9th student, self-studying 8th grade plane trigo) lol

[worthy sapphire]

law of sines is cool

[worthy sapphire]

this is false though, i did the rest in my head apparently

[hot turtle]

What was ur answer btw

[worthy sapphire]

the method i described gives an answer

i'm more interested in the sum being constant for all z1 and z2 rather than the sum itself though...

[smoky jetty]

lol can u give a law of sine problem? maybe i could refresh my knowledge of it. been a while since i did some alot

[worthy sapphire]

Sure!

Let ABC be an acute triangle and let X be a point in its interior. Let D,E,F be the feet of X onto BC,CA,AB. Suppose that DEF~ABC. Express the area of DEF in terms of the area of ABC.

[pulsar grove]

Just go to the help section

[sick hemlock]

It's FKG I think

I mean I see like 9 points that are collinear

But the options limit them

[smoky jetty]

uhh im having a hard time visualizing this

[sinful tusk]

The regions ABC and CDE have the same area. Find the value of α, correct to 2 decimal places.

[plucky mantle]

Is that

A 30 60 90 triangle

Idk much about geometry so that probably isn't it

[pallid quest]

Where are the asymptotes in tangent and cotangent graphs?

[upper karma]

@pallid quest you can also ping the "<@&286206848099549185>" people if you need help.

[pallid quest]

ok thx

[upper karma]

and I just accidentlly pinged numerous of people wow

[pallid quest]

<@&286206848099549185>

[upper karma]

inclduing my self.

[lament smelt]

helper role is basically "discord ping simulator"

[pallid quest]

is there a physics section btw

[upper karma]

nah

[pallid quest]

can you help tho

[lament smelt]

idk

[pallid quest]

bro

[hard oxide]

Any point of the form pi/2 (2n+1)

Where n is an integer

[pallid quest]

Uh

That seems stinky

What about the unit circle?

[hard oxide]

So basically πk/2 for any odd integer k

[hard oxide]

What u mean

[pallid quest]

Dodson said how u can use x/y for the unit circle

[nocturne remnant]

when the point is on the uppermost and lowermost point of the unit circle the corresponding tangent is undefined

[hard oxide]

Whenever there is a straight line

But yea what Elon lauki said

Musk

Mask

[nocturne remnant]

lol whattttt

[hard oxide]

Idek

[pallid quest]

So basically when it’s 0?

[nocturne remnant]

for cotangent its when the point is the rightmost or leftmost, then x/y is undefined

[hard oxide]

Ye

[slate mauve]

how would I solve for G? I was told to make a parallel line in between both lines to find exterior angles but i dont quite understand. some help would be much appreciated. https://media.discordapp.net/attachments/426033723448885248/1019064280710139974/IMG_4019.jpg?width=683&height=910

[restive abyss]

I'm having trouble finding the coords to the midpoint of a line segment given just the ratio and the terminal and intial coords

[sinful tusk]

^

[wraith dawn]

geometry boys

any online?

[sick hemlock]

given a ratio of what exactly?

you could just put terminal and initial coords into the midpoint formula and get the answer

unless I'm misunderstanding the question

[upper karma]

I got a lot questions for homework with this style where they request two values, cos and tan for an angle with a sine condition. I tried to solve them, but it was incorrect according to the answer sheet (doesn't show how).

Calculate the exact values of the cosine and tangents of the angle α, when sin α = -(3/11) and -(π/2) ≤ α ≤ 0.

[dark sparrow]

is this one of the questions you attempted to solve? if so, then show your work so we can pick it apart.

[upper karma]

sorry for late reply

[dark sparrow]

(3/11)^2 is certainly not 112/121...

[slate mauve]

@frigid lantern

[frigid lantern]

Find an equation for the perpendicular bisector of the line segment whose endpoints are (1,-6) and (-7,6)

I need some assistance

[restive abyss]

nevermind, our teacher gave us the wrong homework lol

we didn't learn the formula yet

[upper karma]

it would be 9/121

[north vigil]

I think g is 84° if the pink line and the line that 56° is sitting on are parallel

(If theyre not parallel then i have no clue)

[north vigil]

Basically, if you do 180° - 144°, you will get 36°, that has to be the upper angle on g, now 56 is the lower angle because it tells you, so 56° + 36° would be 84°

[sick hemlock]

It's basically just section formula for 1:1 ratio. Say,
Your initial coords are: (x1, y1)
Your terminal coords are: (x2, y2)
Some point(x, y) divides them in the ratio m:n
x = (mx2 + nx1)/(m+n)
y = (my2 + ny1)/(m+n)

[sick hemlock]

Every point on the perpendicular bisector is equidistant from the endpoints

so you could assume a single point on the bisector as (x,y) and use distance formula

I'm currently outside so I can't do it on paper

[turbid quail]

hey does anyone have the trigonometry formulas. A and B ones

identities

[dense bough]

which ones, like $\sin(A + B) = \sin A \cos B + \sin B \cos A$?

[somber coyote]

findingsouth

[dense bough]

then search up "angle addition/subtraction formulas"

or just "sin(A - B)" et.c

[turbid quail]

ok

[dense bough]

cause there's more than one

there are also formulas for sin A sin B etc. here: https://en.wikipedia.org/wiki/Prosthaphaeresis

[glass jasper]

Help finding the area of the black square.
The equations of the lines are y=sqrt(3)x and y= (1/(sqrt(3)))x. The vertical line has the equation x = 4

[glass jasper]

tan?

[dark sparrow]

i don't understand your question.

[glass jasper]

Like do I use one of the trig functions

[dark sparrow]

just one? no, surely not.

[cedar flower]

Which one, among the following two, is the more common / more natural definition for a line to be parallel to a plane in R^3?

A line is parallel to a plane if they have no intersection

A line is parallel to a plane if its direction is orthogonal to the orthogonal direction of the plane

The two definitions are not equivalent, because they disagree on whether or not lines contained in a plane are parallel to it

[quiet blade]

Wiki says the first one https://en.wikipedia.org/wiki/Parallel_(geometry)#A_line_and_a_plane

[brittle rivet]

Can someone help me with number 67

[sick hemlock]

I think it's 150° or 210°

I need to go check though

It's 210° according to the wiki

[sick hemlock]

Since cosec 210° is cosec (90×2 + 30), it's value should be 2. For cosec to have a negative value, the ray should lie in the third quadrant

[waxen lotus]

When you rationalize a root, does the number become negative root 16 and root 7 or does it combine?

[brittle rivet]

It was 210 and 330

[brittle rivet]

The answer on the top will be negative but you're skipping a step. Start by undoing root -16. It's a perfect square which is 4. Then you can multiply the numerator and denominator by root 7.

[waxen lotus]

Ah, yeah. True. Forgot about that bit

[brittle rivet]

This would be your answer

Squaring the bottom by root 7 just undoes the square root. And multiplying -4 by root7 just gives you -4root7

[sick hemlock]

330? Shouldn't it be 300?

You're right

It's 330

Nvm

[sand citrus]

A graphical approach to algebra and trigonometry by john hornsby, margaret lial, gary rockswold
is this a good book to learn trigonometry, i am currently in grade 12

[brittle rivet]

I would say any book outside your current text is probably good. It's always good imo to have multiple sources for math. Though most books are written in mathanese (heavy mathematical language) so be weary about terms and stuff. When we describe stuff in mathematics we use precise terms to limit any ambiguity. Most books won't stray too far from this.

[brittle rivet]

Well my books answers are 330 and 210. Idk WHY tho.

[slim flax]

Anybody know how to solve?

[sick hemlock]

No I was wrong, 330 is correct

[brittle rivet]

I asked my teacher today and I was overthinking the shit out of it xD

[sick hemlock]

what did your teacher say?

I mean there is a method to this

[brittle rivet]

Well what's the inverse of csc?

Sine

[sick hemlock]

yeah

[brittle rivet]

So csc=-2 is just sin=-1/2

[sick hemlock]

yep

[brittle rivet]

You know why is neg so it has to be in quad 3 and 4

[sick hemlock]

yeah

[brittle rivet]

Then you know sin 1/2 builds your 30 60 90 triangle

360-30=330 and 180+30=210

Bada bing

[sick hemlock]

Yeah, but shouldn't there be more values than just these 2?

[brittle rivet]

What is sin 1/2 my guy

[sick hemlock]

For example csc 570°

[sick hemlock]

sin30 is 1/2 if that's what you're asking

[brittle rivet]

Yes exactly

You don't need anymore

You have your triangle

Sin is r/Y

[sick hemlock]

still, what should be the value of csc570° according to you?

Maybe I'm just making silly mistakes rn but it's -2 according to me

[brittle rivet]

Hold up gimme a sec

[sick hemlock]

Oh yeah

I didn't read the damn question

It's asking for angles till 360°

[brittle rivet]

Right

He talked about how they should have really ended the [0,360) as [0,360]

]this means it stops here

)this means it can go past

The notation was incorrect

[sick hemlock]

yeah I have no idea what these different notations mean tbh

I'm still pretty new to this

[brittle rivet]

Well I'm 27 and took this when I was like 16 so I'm 11 years late to this lol

[sick hemlock]

eh, never too late for maths

[brittle rivet]

But from what I understand is []brackets mean exactly in between these numbers and ()can go outside of those numbers

[sick hemlock]

alright, thanks

[brittle rivet]

So when you say [0,360) it means start at 0 no further back until 360. But if your answer goes past so be it. If it is [0,360] it means start at 0 got to 360 and no further.

[sick hemlock]

oh

[brittle rivet]

[=stop sign . (=some suggested spot

Might want to Google to clarify but that's how I understood it

[sick hemlock]

well thanks for telling me, learnt something new today even though it's well past midnight

[brittle rivet]

Np! Get some sleep!

[sick hemlock]

yeah, goodnight

[carmine oasis]

hi

[hollow onyx]

Hi

[carmine oasis]

do you guys do walk thrus ?

[hollow onyx]

Like for subjects or a question

If you have a certain question you don’t understand someone can walk you through solving it

[brittle rivet]

If you have a question I would start with the ?how-to-get-help section and post your question. You'll get a faster response. You can ask here and maybe someone will help you. It's not really guaranteed.

[carmine oasis]

got it

[oak eagle]

pls help
Find x if the sequence 5, 10, x+2 is a (arithmetic) and (b) geometric

[dark sparrow]

@oak eagle do you still need help with this

[sick hemlock]

Probably does considering it's only been 10 minutes

[dark sparrow]

they might have figured it out on their own

if not then they would do good to respond anyway

[sick hemlock]

fair

[polar urchin]

Hey, Ive been struggling with this one for hours.

Find the second-smallest positive solution of 7sin(x+5)=1.

Ive isolated x down to
x=arcsin(1/7)-5

But am confused on how to get the multiple solutions. I thought you just add 2pi to the end of the equation to get the next solution but that is wrong. I'm stumped!

[grave pond]

arcsin(1/7)-5 is not positive in the first place.

[polar urchin]

@grave pond What other way could I isolate X?

unless its undefined, but the question does not allow undefined as an answer

[grave pond]

You also need to take into account that sin(pi-x)=sin(x), and arcsin will only give you the solution that lies on the upslope ...

[polar urchin]

I'm not totally understanding. For context I'm almost 10 years since my last math class of geometry and teaching my self trig!

[grave pond]

Have you made a plot of the function to help you get an overview yet?

[polar urchin]

I'm trying to. But I'm stuck on simple graphs plugging in 1...2...etc for X and plotting, but I dont know what numbers to plug into this equation

[grave pond]

https://www.desmos.com/calculator/upxgorzwy8

[polar urchin]

So say I'm looking at this graph, I dont quite understand what it means when it says looking for the smallest or second smallest solution

[grave pond]

You can see on the graph there's a solution around -5. That is not positive.

There's a solution between -2 and -1. That is not positive either.

There's a solution beween 1 and 2. That is positive, clearly the smallest positive one.

There's a solution between 4 and 5. That is also positive, and it's the second smallest of the positive solutions, so it's the one you're looking for.

[polar urchin]

In exact form, how do I write that in an equation

it wants x isolated, so I'm still stuck on trying to simplify the problem down to a form that returns the point between 4-5

[grave pond]

One way to proceed if you don't see what it is, would be to write a table of arcsin(1/7)-5+k·2pi and (pi-arcsin(1/7))-5+k·2pi for various integer values of k and see which of them is the one between 4 and 5.

[hazy basalt]

Someone help me on this please

Can anyone help me with this please

I need someone’s help

I don’t got much time

[abstract torrent]

These are the answer from 18 to 20

[unkempt zodiac]

x=80, just i need help with explanation of how angle x is twice angle c

Angle C is 40, Z=50-40 since ABC is isosceles

But how is C 40 thats all i need to know

I know that the center is tiwce at the circumference but how can we know C is the circumference

[sinful remnant]

So what is the given information then? It could be easier to help if we knew the given info by itself and what you are trying to find. I gather you are trying to find the value of angle x

[unkempt zodiac]

x and z

ive found them

x 80

z 10

I gave all the info alread

[upper karma]

im a lil slow

can someone gimme a step to step guide

[upper karma]

nvm

i gotit

[floral shale]

There was a problem just like this

Which yes

x is 80 z is 10

[upper quarry]

is there any difference with sin/cos graphs

im talking about their vertical shifts

y=3cos(x+pi/8)

would it not just be -pi/8

[floral shale]

$\sin\left(x + \frac{π}{2}\right) = \cos(x)$

[somber coyote]

Umbraleviathan

[upper quarry]

https://gyazo.com/44e3e3a81e110ce60cfac1d563c32af3

would this still work in relation to this question

in finding amplitude, horizontal and vertical shifts and period

[floral shale]

no

you would just have to analyze the function

and then graph it based on that analysis

[near harness]

C is the circumferential angle because the vertex of angle C is on the circumference of the circle.

In other words, it is the circumferential angle because point C is on the circumference.

Angle O is the central angle because point O lies on the centre of the circle.

[hollow onyx]

how did this happen

i solved it by adding one to both sides then dividing by two then rooting both sides and i got x = pi/4 (answer group is [0, pi/2]) is this correct?

[quiet blade]

yeah that's valid

on a side note

$0 = 2\cos^2 x - 1 = \cos^2 x + \cos^2 x - 1 = \cos^2 x - (1 - \cos^2 x) = \cos^2 x - \sin^2 x =\cos(2x)$

[somber coyote]

peaceGiant

[hollow onyx]

thanks

[hollow onyx]

wait i don't understand how the 1 went away

[quiet blade]

1 - cos^2 x = sin^2 x; follows directly from the famous sin^2 x + cos^2 x = 1

[frigid kettle]

Hi all. I watched a video where a teacher asked if we can draw a line intersecting all edges of a random triangle. He accepted the answer where the line goes through a vertex and a point between other vertices, but didn't accept the one where the line goes through an edge. He should have accepted the second one as well because it's not only a point that could be the result of an intersection, right?

[quiet vine]

yo guys

can anybody help me memorize the unit circle

i have a test tmr

and i need to memorize it

[safe schooner]

does anyone know how the answer became 225/2?

this is converting radius to degree btw

[silent plank]

the work is pretty much all there,
cancel common factors etc, basic fraction simplification

[silent plank]

do you know the first quadrant

sin,cos,tan of 0,30,45,60,90?

[quiet vine]

nope

[dark sparrow]

there's a mnemonic based on the fingers of a hand

[near harness]

There are several methods. Here's the one i use.

[Order: 0°, 30°, 45°, 60°, 90°]
• For sin: 0, 1/2, 1/√2, √3/2, 1
• Just do the complete reverse for cos: 1, √3/2, 1/√2, 1/2, 0
For finding tan, just do sin/cos

[near harness]

π and π get cancelled.
• 180/8 => divide both numerator and denominator by 4 => 45/2

• 45*5/2 = 225/2

[raven wolf]

radian, and pi radian is 180°

[safe schooner]

ah thanks dudes 👍

[wispy fog]

help plz

[dark sparrow]

bad steps

was better off finding the general value for 2x + pi/6 and then solving for x from there

[wispy fog]

ah rip

wdym by gen value of 2x +pi/6

[dark sparrow]

like if you were asked to solve this over the whole number line rather than that specific interval.

[floral shale]

arcsine is a lifesaver

[near harness]

$\sin{2x+\frac{\pi}{6}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

\sin{\frac{\pi}{4}} = \frac{1}{\sqrt{2}

2x = \frac{\pi}{12}

x = \frac{\pi}{24}

OR x = \frac{15}{2} degrees$

[somber coyote]

Nishil

$\sin{2x+\frac{\pi}{6}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

\sin{\frac{\pi}{4}} = \frac{1}{\sqrt{2}

2x = \frac{\pi}{12}

x = \frac{\pi}{24}

OR x = \frac{15}{2} degrees$
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.58 
     
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```

[near harness]

I'm new to LaTeX so idk why it's showing this error

[wispy fog]

got it thanks

[near harness]

Most welcome

[wispy fog]

would this be 3

i think its 3 or rt3 but idk tbh

[hollow onyx]

anyone have a pdf that has all trigonometric identities even derived ones

[grave pond]

All trigonometric identities? That would make the list infinitely long.

[hollow onyx]

like something that has all of them that i need i don't think i need an infinite amount of trig identities

[floral shale]

1 + 1 = 2

If you let 1 = sin^2(x) + cos^2(x)

[calm umbra]

This has a lot of the common ones you'd learn/use in pre calc or calculus

Which i found by going to google images and typing "trig identities"

A lot of math formulas/theorems/equations etc are easily found by typing what you want on google images

[near harness]

I think it will be -3. I'm Not sure about the answer as well

I got |x| = √3
So x = ±√3

[wispy fog]

Yep I got the same in the end thanks mate

[near harness]

You're most welcome

@wispy fog so do you have an ebook or something? I was just curious about where you got these problems from.

If you don't mind, can i get access to them?

[hollow onyx]

Thanks

[eager quiver]

have no idea what to do

[hollow onyx]

Take the length dc as x

Label the angle near b as $\theta$ or anything you want

[somber coyote]

mstf

[hollow onyx]

Wait do you have a calculator

[eager quiver]

ive done it now

stuck on this question

[heavy shale]

can u use calculator

[eager quiver]

yes

[heavy shale]

then do tan 41 = bd/cd

and find bd

use bd/ ab = tan $\theta$ where $\theta$ = angle A

[somber coyote]

NotMyself

[heavy shale]

find theta

then cos $\theta$ = ab/ ac

[somber coyote]

NotMyself

[heavy shale]

or u can use Pythagoras $((bd)^2 + (ab)^2)^(1/2)$

[somber coyote]

NotMyself

[hollow onyx]

do $a^{1/2}$

[somber coyote]

mstf

[hollow vortex]

On this problem right here I dont understand why sina = 12/180. since sine is opp/hyp and a does not even face 12.

https://cdn.discordapp.com/attachments/326138772477575180/1021542260984266833/IMG_0304.jpg

[vestal chasm]

why do u convert rev to rad lol arent they the same meaning because one full revolution is a full rotation or 2pi why divide by 2pi if its already a ful rotation already

like isnt revolution and rotation synonyms

this is the correct answer ^ what the professor said

[hollow vortex]

can someone please check

[digital halo]

If the upper side sin a is parallel to the line Chicago/Saint Louis, a is the angle on Chicago

[hollow vortex]

by upper side you mean the line above angle a?

[digital halo]

Round its 8 o'clock

[dire chasm]

Can anyone help me with this?

[hollow vortex]

oh your saying if there was a line from st louis to a then it would be 12 but wouldent that be adjacent?

[digital halo]

1mn, I know the name of the inside/outside angle in French, I'm looking for the correct traduction in English.

[hollow vortex]

k

[digital halo]

Alternates angles. If 2 parallel lines are crossed by a third lune, the angles at each parallel lines are equal.

Does in your problem it says, that Chicago/Saint Louis line and line passing by Position point are parallel? Or same azimuth?

Because ot is one way to say angle a equal angle at Chicago.

Then you can use the sin-1 (12/180) relation.

Also does your problem is in Euclidean or Spherical Geometry?

[hollow vortex]

I think Euclidean and it dosent say if its parrell or not it litterally just did those calculations without explaining any of the logic

[digital halo]

Sorry working now

[hollow vortex]

ok

[digital halo]

Ok I got some time, I will look a little bit more seriously.

Ok I read all, It is as I told you. The line from position which point at c, is parallel to the line Chicago Saint Louis.

Did you try in help section, first?

[ivory patio]

Oops

Sorry

[digital halo]

So , because the 2 lines are parallel

you can make some deduction about angle

and as reply by image before, the alternates angles rules applies, That where he got sin-1(12/180)

and he same goes for b angle with the angle from Saint Louis Looking at 12.

To all and anybody. I have a question about plan equation and sphere. I got the solution all figured, but I got it on a hunch, and I would like to dig the why it works.

Here is the math of it, Just after I will explain how I use it.

So, I create for fun a little spreadsheet to do some celestial navigation. To do that, I work on a sphere of radius 1. I need to find the the intersection of 2 plans with the sphere.

Hence the formula search above. As you see it is based on plan equation of the format ax+by+cz+d=0 and a sphere equation where the 3D Pythagoras is used.

[honest elm]

There’s a 3d pythagoras?

[digital halo]

The problem I got was when I was calculating the discriminant.

In Mechanic and Piping it is known. Distances between 2 points A(xa;ya;za) and B(xb;yb;zb), the distance d between the 2 point is Sqrt((xb-xa)^2+(yb-ya)^2+(zb-za)^2)

[stone heron]

the moment you said that, it reminded me of the unit vector calculations and the whole sqr (x^2+y^2+z^2)

I just remember drawing 2 2d triangles to make a 3d triangle and subbing in stuff to get that

[digital halo]

So my discriminant. with many set of data, my discriminant turn ou to be negative, but that was impossible for me becaus eI can see and mesure the solution, it was not an imaginary number i.

So I got the idea to scale my to plan equation to the same d. because any plane equation on the form kax+kby+kcz+kd=0 whatever is k, give you the same plan.

I got less negative discriminant. But still It was not an enough, I wanted to work in any case.

That's when I go my second hunch, and scale both of the plan equation in order that any of their argument the bigger one is 1 max.

And it works flawlessy eversince.

I see why I have to scale, because it force discriminant to be positive. But what is this goddamn d, how I can visualized it? Why scalling my equation to got the same d in each plan equation? It is like I'm bringing the 2 plans in the same dimension! Like the factor k is a gate to imaginary dimension.

So any help to help understand why my hunches worked, will be appreciated, it is all time in a corner of my mind!

Any taker?

[digital halo]

In addition , like a sphere. The square radius is the sum of the square of its x y and z difference with the center.

[hasty karma]

cos(a+-b)=cos a cos b -+ sin a sin b

cos( 7pi/15 + 11pi/30) = cos((14pi+11pi)/30) = cos(25pi/30) = cos(5pi/6)

5pi/6 is ~150 deg

180-150 = 30, cos = -(sqrt 3)/2

simple one

[heavy shale]

$cos (A \pm B) = cosAcosB \mp sinAsinB$

[somber coyote]

NotMyself

[heavy shale]

the formula used^^

[hasty karma]

same thing